The running engine of an automobile produces carbon monoxide (CO), a toxic gas, at the rate of about .102mol CO per minute. a car is left idiling in a poorly ventilated garage that is 60m long, 4 m wide, and 2.2 m high at 20 degrees C. what will the partial pressure of CO in one minute?
Why don't you limit your self to asking 1-3 questions. We don't you your dam homework for you Missy.
Use PV = nRT
Solve for P, you have V(or can calculate it but you want it to be in liters), you have n, R, and T.
To determine the partial pressure of CO in the garage after one minute, we need to calculate the number of moles of CO that will be present and then divide it by the volume of the garage.
First, let's calculate the volume of the garage:
Volume = Length × Width × Height
Volume = 60 m × 4 m × 2.2 m
Volume = 528 m^3
Next, let's calculate the number of moles of CO produced in one minute:
Number of moles of CO = Rate of CO production × Time
Number of moles of CO = 0.102 mol CO/min × 1 min
Number of moles of CO = 0.102 mol
Now, we can calculate the partial pressure of CO using the ideal gas law:
PV = nRT
Where:
P = Partial pressure of CO (unknown)
V = Volume of the garage (528 m^3)
n = Number of moles of CO (0.102 mol)
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
To convert the temperature from Celsius to Kelvin, we need to add 273.15:
T = 20°C + 273.15 = 293.15 K
The ideal gas law becomes:
P × 528 m^3 = 0.102 mol × 8.314 J/(mol·K) × 293.15 K
Rearranging the equation to solve for P:
P = (0.102 mol × 8.314 J/(mol·K) × 293.15 K) / 528 m^3
P ≈ 0.144 Pa
Therefore, the partial pressure of CO in the garage after one minute is approximately 0.144 Pascal (Pa).